Transcript March 7, 2011

March 7, 2011
Special Relativity
Reading: Chapter 4, Rybicki & Lightman
PRELIMINARIES
History
> SR developed out of a need to reconcile classical, Newtonian mechanics,
Maxwell’s Equations and the laws of EM
> SR does not deal with gravitation, only electromagnetic forces
Inertial Coordinate Systems
A key concept in SR is the definition of inertial reference frames:
A reference frame is inertial if every particle within it which is
initially at rest remains at rest, and every test particle in motion
continues that motion without change in speed or direction.
Then in SR 
All the laws of physics are the same in every inertial inertia frame.
“Principle of Relativity”
Example:
An inertial reference frame is a space ship in free fall near the Earth
• A free particle at rest in the vehicle remains at rest
• A particle given a gentle push moves across the vehicle in a straight
line with constant speed.
Why? Because the particle and the ship are both falling with the
same acceleration towards the Earth
• The space ship is only an inertial reference system when it is
small enough that differences in the accelerations over its size can be
ignored.
• If particles didn’t have the same acceleration regardless of size,
shape, composition, etc. then an observer inside the space ship
would notice a relative acceleration among different particles.
 some of the particles at rest would not remain at rest, others would.
A sufficiently large space ship would not be an inertial reference frame,
because the acceleration by gravity is not uniform across it.
from Spacetime Physics by Taylor & Wheeler
“The laws of physics are the same
in every inertial reference frame”
• This means that if you derive a law of physics in one frame, you can apply
it in another.
• Both the form of the laws of physics and the numerical values of the
physical constants that the laws contain are the same in every frame.
• All inertial frames are equivalent in terms of every law of physics
 the laws of physics cannot provide a way to distinguish one
inertial reference frame from another.
• Although the laws of physics are the same in every frame,
measured quantities (like the time between events A and B, the distance
measured between two points, even the force) will not be the same.
• Only the mathematical form of the laws of physics are the same.
Note: Special relativity can deal with accelerated motion, but
only in inertial frames
Events
The concept of an “event” is crucial in SR.
An “event” is something that “happens” at a specific location
in space (x,y,z) and at a specific time, t
The key question in SR is:
How do the coordinates x,y,z,t of a given event relate to the
coordinates of the same event as measured by an observer
in another inertial coordinate system.
One imagines a lattice-work grid to measure x,y,z and a bunch of
“clocks” which are somehow synchronized by pulses of light, say,
and then can record when a particular “event” happens at point
x,y,z
These things seem intuitively obvious, but what is meant by a
measurement at a point x,y,z,t is crucial.
“The Observer”
The word “observer”
is a shorthand way of
speaking of the
whole collection of
recording clocks
associated with one
inertial frame.
For simplicity, we will define our inertial coordinate systems for
two observers in such a way that
the origins are chosen and the clocks synchronized in such
a way that when the origins coincide, the clocks read zero.
The speed between the two systems is parallel to the x-axis.
y
y'
v
x'
x
z
z'
So we want to connect the coordinates
of some event in one frame (x,y,z,t) to
the coordinates of the same event in
the other frame, (x‘,y‘,z‘,t‘)
x  x  vt
y  y
z  z
Before SR, the transformation was obvious:
The Galilean Transformation
t  t
However, the fact that physical constants are the same in all frames
implies that the SPEED OF LIGHT (c) is the SAME IN ALL FRAMES:
A pulse of light along the x-axis in the “unprimed” system travels
x A  ctA
in time
In the “primed” system, it also travels
But:
xB  x A  vt A
tB  t A
tA
xB  ctB
since light travels
at c in all frames
So the Galilean transformation predicts
xB  xA  vt B  ctB
Instead, one can show that the proper transformation is
the Lorentz
Transformation.
x   ( x  v t )
y  y
  1
where
or letting
z  z
t    (t  cv2 x )


 
v2
c2

1/ 2
v
c
1
1  2
The reasons why the transformation takes this form are discussed in
texts about SR, but note that since
t  t
we do satisfy the “pulse of light” argument, since
x y z c t
2
2
2
2 2
2
2
2
2 2



x  y  z  c t
transform correctly
The inverse of
x   ( x  v t )
y  y
is
z  z
t    (t  cv2 x )
x   ( x  v t )
y  y
z  z
t   (t   cv2 x)
which looks like the original, except that “primed” and “unprimed”
variables are interchanged, and v is replaced by -v
Note:
deviations from Galilean transformations become large when
v
~1
c
Our every-day experience involves
v  c
v ~ 10-6 c in a jet plane
v ~ 2x10-5 c in a satellite
Some implications of what this means:
K frame: unprimed
K’ frame: primed
Length Contraction


L0  x2  x1
Consider a rigid rod of length
at rest in frame K’
L  x2  x1
What is the length in frame K?
where x2 and x1 are the positions of the ends of the rod
measured at the same time t in the K frame.


L0  x2  x1
  ( x2  x1 )

  L
L  1
The rod appears shorter by a factor

1


 1
v2
c2

v2
c2
1/ 2

1/ 2
L0
This is symmetric:
K sees a rod which appears shorter by a factor of 1/γ
K’ would see the same rod held up by an observer in the K frame
contracted by 1/γ
This is because the observers in the 2 frames would not agree that
the ends of the stick were measured at the same time
in both frames.
Time Dilation
Suppose a clock at rest at the origin of the K’ frame
measures an interval of time


T0  t 2  t1
What time interval does an observer in frame K measure?
In K’ we have x’=0, so the time interval measured in K is
T  t 2  t1
 
  (t 2  t1 )
T   T0
The moving clock appears to have slowed

1
1  2
down.
Similarly, K’ will think K’s clocks have slowed down.
Relativity of Simultaneity
Events which are simultaneous in one frame are NOT simultaneous
in any other frame.
All the clocks in your inertial frame seem to be going at the same rate,
but clocks in the moving frame differ from one another depending
on their LOCATION.


t   t  cv2 x
So at a given t in the K frame,
t  cv2 x
must be some fixed
value for all clocks in the K’ frame

Thus the greater x’ is, the smaller t’ is.
This effect is the explanation for most “paradoxes” in SR.
To illustrate, Imagine we have two clocks in the K’ frame at positions
A and B, separated by length L’.
A flashbulb, located exactly in the middle, goes off.
A and B are instructed to set their clocks to t’=0 when the flash
reaches them, which in the K’ frame, happens at exactly the same time
L’
K’
What does K see?
What does K see?
At t=0, the flash goes off, and the
wavefront starts to expand
K
v
v
t=0
B
A
At t=tA, it reaches Point A, and A
resets his clock at t’ = 0
t=tA
A
Sometime later, at t=tB, the
wavefront reaches B, and he resets
his clock, also at t’ = 0
Thus, to an observer in the K frame,
A has reset his clock BEFORE B reset his clock....
whereas in K’, the clocks were reset SIMULTANEOUSLY.
B
t=tB
A
B
Cosmic-Ray Mesons: Observed time dilation
Bruno Rossi & D.B. Hall 1941, Phys Rev, 59, 223
Cosmic Rays impinging on the top of the Earth’s atmosphere produce
μ-mesons which decay via
   e 1  2
with some half-life.
neutrinos
• The mesons travel down through the Earth’s atmosphere at v ~ c
• From the half-life for decay, one can estimate how many should be
seen at sea-level vs. high in the atmosphere, given the travel time.
• Many more are seen at sea level than expected
 their radioactive “clock” appears to be running slow.
GPS: Observed time dilation
• GPS system consists of a network of 24 satellites in high orbits around
the Earth.
• Each satellite has an orbital period of about 12 hours, and an orbital
speed of about 14,000 km/hour
• The satellite orbits are arranged so that at least 4 and sometimes as
many as 12 satellites are visible from any point on Earth.
• Each satellite carries an atomic clock which “ticks” with an accuracy of
1 nanosecond. (Actually they each carry 2 cesium clocks and 2
rubidium clocks).
• A GPS receiver determines your position by comparing time of arrival
signals from a number of different satellites, and thus figuring
out how far you are from each satellite -- “triangulation.”
• The inexpensive hand-held GPS receivers can determine your position
to 5 or 10 meters. To achieve this accuracy you need to know the
clock ticks to an accuracy of 20-30 nanoseconds. Military applications
and airliners have more accurate GPS receivers.
• Special relativity: the clocks on board the satellites will run slower than
atomic clocks on the ground by 7 microseconds per day because of time
dilation.
• General relativity: The clocks on board the satellites run FASTER than
atomic clocks on the ground by 45 microseconds a day because the
curvature of space-time from the Earth’s gravitational field is LESS
farther from the surface.
• If you don’t take these two effects into account then you would
be as much as 10 km off from the build up of errors each DAY.
• To solve this basically the number of cesium atom “ticks” per second is
redefined for the clocks on the satellites so that “one second” on the satellite
is the same as “one second” on the ground.
(on the ground a CS-133 clock frequency is 9,192,631,770 Hz)
TRANSFORMATION OF VELOCITIES
Suppose a particle has velocity

  
u   (u x , u y , u z )

u  (ux , u y , uz )
What is its velocity
y
in frame K?
y’
K’
K
v
x
Lorentz
Transform
x   ( x  v t )
y  y
z  z
t    (t  cv2 x )
x’
in frame K’
Differentials 
dx   (dx  v dt )
dy  dy
dz  dz 
dt   (dt   cv2 dx)
so...

ux  v
dx  dx  vdt 
ux 



dx  
dt
 
  dt  v 2  1  vu x
c 

c2
uy 
uy

 vu  
 1  2x 

c 


uz 
uz

 vu  
 1  2x 

c 


Note: These formulae can be used to add velocities
e.g. Suppose

u x  0.5c and v  0.5c
In Galilean transforms, you expect

ux  ux  v  c
But, using the Lorentz-transform formulae
0.5c  0.5c 4
ux 
 c c
2
1  0.5
5
sums of velocities are ALWAYS < c
TRANSFORMATION OF ACCELERATIONS
Similarly, one can ask how accelerations transform from K to K’
velocity
ux 

u

ux  v
1
v u x
c2
so





 u   v v du  
dux
x
x 
dux 
 
2
2

c





vu x
v
u
1 2
 1  x 

2
c


c 



2

1  v c2 


du
x
2



v
u
x
1 

2

c 



dux

2


v u x 
2
 1 2

c 


Recall
So



ux 
dx 



dt    dt  v 2    1  v 2 dt 

c 
c 



dux
ax 

dt
du x
dt 
 vu  
3
 1  2x 

c 


ax 
ax
3

 vu  
3
 1  2x 

c 


3
Similarly, can show
vu   
 y
a
2
x



c
ay


ay 

2
3


 vu 
 vu 
 2 1  2x 
 2 1  2x 


c 
c 




So ax and ay are related to ax’ and ay’ by some fairly complicated
expressions involving not only v, but also ux’ and uy’
If a body is instantaneously at rest in the K’ frame, then
ux’ = uy’ = 0 and
ax is diminished by γ3
ay is diminished by γ2
Transformation of Angles
(Aberration of Star Light)
We saw how ux, uy, uz transformed to ux’, uy’, u’z
for v = relatively velocity of frames along x-axis.
We could also write u in terms of the components parallel and
perpendicular to v, where v is in some arbitrary direction with
respect to x,y

u||  v
u|| 

vu||
1 2
c
Then
y’
y
u
u

u||


 vu  
 1  2|| 
c


u||  u cos
u’
x
u 
u
x’
u  u sin 
The angle
vu||
1 2
u
u
c
tan  

 u   v

u||
vu||  ||

 1 2 
 c 
u

 u||  v
 
usin 

 ucos v
The interesting case is u=u’=c then

where
t an 

u  u 
sin  
v

  cos   
c

Or, in terms of cosθ:
ux  c cos
so
cos 
ux
c

also
ux  v
ux 

vu x
1 2
c

1 ux  v


c
v ux
1 2
c
1 c cos   v

c 1  ccos  v
c2
cos   
cos 
1   cos 
v

c
Stellar Aberration
Discovered by James Bradley in 1728
Bradley was trying to confirm a claim
of the detection of stellar parallax,
by Hooke, about 50 years earlier
Parallax was reliably measured
for the first time by
Friedrich Wilhelm Bessel in 1838
Refn:
A. Stewart: The Discovery of
Stellar Aberration, Scientific American,
March 1964
Term paper by Vernon Dunlap, 2005
Because of the Earth’s motion in its orbit around the Sun, the angle at
which you must point a telescope at a star changes
A stationary telescope
Telescope moving at velocity v
Analogy of running in the rain
As the Earth moves around the Sun, it carries us through a
succession of reference frames, each of which is an inertial reference
frame for a short period of time.
Bradley’s Telescope
With Samuel Molyneux, Bradley had master
clockmaker George Graham (1675 – 1751) build
a transit telescope with a micrometer which
allowed Bradley to line up a star with
cross-hairs and measure its position WRT zenith
to an accuracy of 0.25 arcsec.
Note parallax for the nearest stars is
~ 1 arcsec or less, so he would not have been
able to measure parallax.
Bradley chose a star near the zenith to minimize
the effects of atmospheric refraction.
.
The first telescope was over 2 stories high,
attached to his chimney, for stability. He later
made a more accurate telescope at his
Aunt’s house. This telescope is now in the
Greenwich Observatory museum.
Bradley reported his results by writing a letter to the
Astronomer Royal, Edmund Halley.
Later, Brandley became the 3rd Astronomer Royal.
Vern Dunlap sent this picture from the Greenwich Observatory:
Bradley’s micrometer
In 1727-1728 Bradley measured the star gamma-Draconis.
Note scale
Is ~40 arcsec reasonable?
The orbital velocity of the Earth is about v = 30 km/s
v
   10  4
c
Aberration formula:
cos  
cos ' 
1   cos
 (cos   )(1   cos )
 cos     cos2    2 cos
(small β)
2

cos  cos   sin 
(1)
Let
    
Then
  angle of aberration
cos   cos(   )
 cos cos  sin  sin 
α is very small, so cosα~1, sinα~α, so
cos    cos    sin 
Compare to (1):
we get
(2)
2

cos  cos   sin 
   sin 
Since β~10^4 radians  40 arcsec
at most
BEAMING
Another very important implication of the aberration formula is
relativistic beaming
sin  
tan 
 cos    
cos  
cos ' 
1   cos
Suppose
Then
   2
tan 
1

That is, consider a photon emitted at
right angles to v in the K’ frame.
sin  
For   1, sin  is small
1


1

So if you have photons being emitted isotropically in the source
frame, they appear concentrated in the forward direction.